blair
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Posts: 5
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Post by blair on Jun 10, 2020 14:07:06 GMT
I am trying to understand the justification for the following:
"If the height is y_old at some point x, then moving x an infinitesimal distance delta to the right yields a new height given by y_{new} = (1 + \delta)y_{old}."
The way the diagram is drawn, it appears that the triangles with sides y_old and y_new are similar triangles, thus the sides have proportional lengths. But they are not similar triangles, because the bottom side has a fixed length of one. This kind of logic would appear to work the same way for any curve, not specifically e^x.
If similar triangles are not assumed (it was not explicitly said), then I don't know where the result comes from. That is too bad, because it looks like such an elegant derivation.
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Post by Admin on Jun 10, 2020 15:42:55 GMT
blair
The triangles can be said to be ultimately similar: that is, as $\delta$ approaches zero the triangles are similar.
Another way to look at this is to say that if $f'$ is the derivative of $f$ with respect to $x$ then by definition
$\displaystyle f'=\frac{(y_{new}-y_{old})}{\delta}=y_{old}$ where $\delta$ is infinitesimal. Remember that $f'(x)=f(x)$ in this particular case.
It follows from this that ultimately $y_{new}=(1+\delta)y_{old}$.
I hope this helps. Please come back with more questions if you need to.
Vasco
PS, see next post below.
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Post by Admin on Jun 10, 2020 18:39:43 GMT
blair
Note the explanation of infinitesimal starting on page 20 with "We could find $V$..." and continuing onto page 21 and ending with "..., so are $V\delta$ and $M$."
Vasco
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Post by Admin on Jun 10, 2020 19:18:47 GMT
blair
Another way of looking at the similar triangles approach: we note that the small right-angled triangle whose hypotenuse is the line between the two white dots, and the shaded triangle, are ultimately similar. By ultimately we mean as $\delta$ approaches zero.
This then leads to the same result.
This approach is used on pages 21 and 22 using figure 16.
Vasco
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blair
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Post by blair on Jun 11, 2020 15:49:16 GMT
Thank you. This approach produces similar triangles only when f'(x)=f(x) and not for any function as I thought. It also depends on the base of the triangle formed by the tangent line and the x-axis having length one. Intuitively I am still a bit confused. You also get similar triangles when f(x) is a straight line, for any size of delta. So I would expect the triangles to become "more" similar as f(x) approaches a straight line, instead of suddenly becoming similar when f(x)=f'(x).
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blair
New Member
Posts: 5
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Post by blair on Jun 12, 2020 19:50:05 GMT
OK, I think I get it. We are not trying to trace the exponential curve, we are only trying to reproduce the equation for the limit, which also falls below the curve (as shown in figure [20b]) for any finite step. I do appreciate you pointing out that the result in [20a] comes from calculating the derivative, which I did not understand from reading the text.
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Post by mondo on Sept 2, 2022 20:09:57 GMT
This one puzzled me as well. While I see $y_{new} = (1+\delta)y_{old}$ I have a problem with the first sentence on this page "Drawing a tangent at an arbitrary point, the base of the shaded triangle is always equal to 1". Is it because $\tan(\theta) = \frac{Y_{old}}{x} = Y_{old}$ where $\theta$ is equal to the angle between a tangent to the function at a point $x$ and positive $X$ axis and hence $x = 1$?
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Post by Admin on Sept 2, 2022 20:28:28 GMT
This one puzzled me as well. While I see $y_{new} = (1+\delta)y_{old}$ I have a problem with the first sentence on this page "Drawing a tangent at an arbitrary point, the base of the shaded triangle is always equal to 1". Is it because $\tan(\theta) = \frac{Y_{old}}{x} = Y_{old}$ where $\theta$ is equal to the angle between a tangent to the function at a point $x$ and positive $X$ axis and hence $x = 1$? Mondo It's because $f'(x)=f(x)$. So yes you have it right. Vasco |
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Post by mondo on Sept 2, 2022 20:52:36 GMT
Thank you Vasco.
One thing I noticed is that tangent (trigonometric) function is rarely used in the definition of derivative and that's why I had to recall this relation from a deep memory.
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